What is (\partial)/(\partial x)(2ycos(9y^2x)) ?

The answer to (\partial)/(\partial x)(2ycos(9y^2x)) is -18y^3sin(9y^2x)

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(\partial)/(\partial x)(2ycos(9y^2x))

\frac{\partial}{\partial x} (2 y cos (9 y^{2} x))

- 18 y^{3} sin (9 y^{2} x)

Solution steps

\frac{\partial}{\partial x} (2 y cos (9 y^{2} x))

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 y \frac{\partial}{\partial x} (cos (9 y^{2} x))

Apply the chain rule:

- sin (9 y^{2} x) \frac{\partial}{\partial x} (9 y^{2} x)

= - sin (9 y^{2} x) \frac{\partial}{\partial x} (9 y^{2} x)

\frac{\partial}{\partial x} (9 y^{2} x) = 9 y^{2}

= 2 y (- sin (9 y^{2} x) \cdot 9 y^{2})

Simplify

2 y (- sin (9 y^{2} x) \cdot 9 y^{2}) : - 18 y^{3} sin (9 y^{2} x)

= - 18 y^{3} sin (9 y^{2} x)

x \to 0 lim (x^{\frac{1}{2}})

\int 2 x tan (x) d x

x \to 2 - lim (4 - x^{2})

n = 0 \sum \infty \frac{6}{n ^{2} - 1}

- 5 t^{3}

What is (\partial)/(\partial x)(2ycos(9y^2x)) ?
The answer to (\partial)/(\partial x)(2ycos(9y^2x)) is -18y^3sin(9y^2x)